Proof Group Homomorphism: $\varphi(e)= e'$ & $\varphi(\hat{a})= \widehat{\varphi(a)}$

Question

I would like to proof the following Lemma:

A group homomorphism $\varphi$ from $\langle G;\ast,\hat{},e \rangle$ to $\langle H;\star,\hat{},e' \rangle$ satisfies:

1)

$\varphi(e)= e'$

Proof: $\varphi(e)=\varphi(e\ast e)=\varphi(e)\star\varphi(e)=e'\star e'=e'$

2)

$\varphi(\hat{a})= \widehat{\varphi(a)}$

Proof: $\varphi(e)=\varphi(\hat{a}*a)=\varphi(\hat{a})\star\varphi(a)=e'$ which requires $\varphi(\hat{a})= \widehat{\varphi(a)}$ from the definition of $e'$

I don't know whether the above proofs are valid. Maybe some improvement on the notation?

Answer 1

1) is wrong because you used the very equality that you are trying to prove, namely at the step $$\varphi(e)\star\varphi(e)=e'\star e'$$

2) is perfect.

A correct way for 1) is to say that $\varphi(e)\star\varphi(e)=\varphi(e)$ implies that $\varphi(e)=e^\prime$ by left- (or right-) cancellation. That is, by multiplying both sides by $\widehat{\varphi(e)}$, whatever that may be. In one line:

$\varphi (e)=\varphi(e)\star \varphi(e)\star \widehat{\varphi (e)}= \varphi (e\star e)\star \widehat{\varphi (e)}= \varphi(e)\star \widehat{\varphi (e)}=e^\prime $

Answer 2

For 1):

Let $x=\varphi(e)$

$$ \varphi(a)=\varphi(ea)=\varphi(e)\varphi(a)=x\varphi(a) $$ Multiply by $\widehat{\varphi(a)}$ on the right, conclude $e'=x$.